Question

what is the positive solution to the equation $0 = -x^2 + 2x + 1$?
quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$\circ -2 + \sqrt{2}$
$\circ 2 - \sqrt{2}$
$\circ 1 + \sqrt{2}$
$\circ -1 + \sqrt{2}$

Explanation:

Step1: Identify a, b, c from the quadratic equation

The quadratic equation is \(0 = -x^{2}+2x + 1\), which can be written in the standard form \(ax^{2}+bx + c = 0\) as \(-x^{2}+2x + 1=0\). So, \(a=- 1\), \(b = 2\), \(c = 1\).

Step2: Substitute into the quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Substitute \(a=-1\), \(b = 2\), \(c = 1\) into it:
\[

$$\begin{align*} x&=\frac{-2\pm\sqrt{2^{2}-4\times(-1)\times1}}{2\times(-1)}\\ &=\frac{-2\pm\sqrt{4 + 4}}{-2}\\ &=\frac{-2\pm\sqrt{8}}{-2}\\ &=\frac{-2\pm2\sqrt{2}}{-2}\\ &=1\mp\sqrt{2} \end{align*}$$

\]

Step3: Find the positive solution

We have two solutions: \(x = 1+\sqrt{2}\) (when we take the minus sign in \(1\mp\sqrt{2}\), wait, no: let's re - check the calculation. Wait, \(\frac{-2\pm2\sqrt{2}}{-2}=\frac{-2(1\mp\sqrt{2})}{-2}=1\mp\sqrt{2}\)? Wait, no, \(\frac{-2 + 2\sqrt{2}}{-2}=\frac{-2(1-\sqrt{2})}{-2}=1-\sqrt{2}\) and \(\frac{-2-2\sqrt{2}}{-2}=\frac{-2(1 + \sqrt{2})}{-2}=1+\sqrt{2}\). Now, \(1-\sqrt{2}\approx1 - 1.414=- 0.414\) (negative), \(1+\sqrt{2}\approx1 + 1.414 = 2.414\) (positive). So the positive solution is \(1+\sqrt{2}\).

Answer:

\(1+\sqrt{2}\) (corresponding to the option: \(1+\sqrt{2}\))